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BQ2004E application note - schematic questions

Other Parts Discussed in Thread: BQ2004E

Hello,

 

I'm using a charge circuit with bq2004e for a 8cell battery pack and i was checking the application note and found some questions,

 

on page 5 from http://focus.ti.com/lit/ug/sluu016a/sluu016a.pdf

 

what is the purpose of the diode 1N4001 (designator D8) and the resistor R13 (2K Ohms). This resistor provides a load current in the battery in any state or case am i right?

other question is why i need a 0R resistor (R10) on the gate of the FET 2N7000.

  • D8 is present to clamp the battery voltage output to no higher than one diode drop above Vin.  Consider the case where the charger is charging at the fast charge current level.  Suddenly, the battery is removed.  The energy/current in the inductor would have to be absorbed by the output capacitor, C1.  This could cause the voltage at BAT+ to go very high.  D8 returns this energy to the input cap, C2, which has sufficient capacity to absorb this energy without a large change in voltage.

    R13 places Vin on BAT+ through a fairly high impedance connection.  The appearance of Vin on BAT+ will force the charger to enter sleep mode when the battery is not present.  Without this resistor, the charger would switch every now and then to get the BAT+ output above its MCV.  The output would then decay back down at which point the IC would turn on again.  R13 keeps the IC in sleep mode.

    R10 can be populated to slow down the turn on of transistor Q2.  This might be needed to reduce EMI.

    Also note that D9 should be drawn as a zener diode.